Solve 1/2-x-1=1/x-2-1/3x^2-12

$$\frac { 1 } { 2 - x } - 1 = \frac { 1 } { x - 2 } - \frac { 1 } { 3 x ^ { 2 } - 12 }$$

Solve for x

$$x=\frac{2\sqrt{3}}{3}-1\approx 0.154700538$$

$$x=-\frac{2\sqrt{3}}{3}-1\approx -2.154700538$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -2,2 since division by zero is not defined. Multiply both sides of the equation by 3\left(x-2\right)\left(x+2\right), the least common multiple of 2-x,x-2,3x^{2}-12.

$$-6-3x+3\left(x-2\right)\left(x+2\right)\left(-1\right)=3x+6-1$$

Multiply 3 and -1 to get -3.

$$-6-3x-3\left(x-2\right)\left(x+2\right)=3x+6-1$$

Use the distributive property to multiply -3 by x-2.

$$-6-3x+\left(-3x+6\right)\left(x+2\right)=3x+6-1$$

Use the distributive property to multiply -3x+6 by x+2 and combine like terms.

$$-6-3x-3x^{2}+12=3x+6-1$$

Add -6 and 12 to get 6.

$$6-3x-3x^{2}=3x+6-1$$

Subtract 1 from 6 to get 5.

$$6-3x-3x^{2}=3x+5$$

Subtract 3x from both sides.

$$6-3x-3x^{2}-3x=5$$

Combine -3x and -3x to get -6x.

$$6-6x-3x^{2}=5$$

Subtract 5 from both sides.

$$6-6x-3x^{2}-5=0$$

Subtract 5 from 6 to get 1.

$$1-6x-3x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-3x^{2}-6x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -6 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-3\right)}}{2\left(-3\right)}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\left(-3\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-6\right)±\sqrt{36+12}}{2\left(-3\right)}$$

Add 36 to 12.

$$x=\frac{-\left(-6\right)±\sqrt{48}}{2\left(-3\right)}$$

Take the square root of 48.

$$x=\frac{-\left(-6\right)±4\sqrt{3}}{2\left(-3\right)}$$

The opposite of -6 is 6.

$$x=\frac{6±4\sqrt{3}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{6±4\sqrt{3}}{-6}$$

Now solve the equation x=\frac{6±4\sqrt{3}}{-6} when ± is plus. Add 6 to 4\sqrt{3}.

$$x=\frac{4\sqrt{3}+6}{-6}$$

Divide 6+4\sqrt{3} by -6.

$$x=-\frac{2\sqrt{3}}{3}-1$$

Now solve the equation x=\frac{6±4\sqrt{3}}{-6} when ± is minus. Subtract 4\sqrt{3} from 6.

$$x=\frac{6-4\sqrt{3}}{-6}$$

Divide 6-4\sqrt{3} by -6.

$$x=\frac{2\sqrt{3}}{3}-1$$

The equation is now solved.

$$x=-\frac{2\sqrt{3}}{3}-1 x=\frac{2\sqrt{3}}{3}-1$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -2,2 since division by zero is not defined. Multiply both sides of the equation by 3\left(x-2\right)\left(x+2\right), the least common multiple of 2-x,x-2,3x^{2}-12.

$$-6-3x+3\left(x-2\right)\left(x+2\right)\left(-1\right)=3x+6-1$$

Multiply 3 and -1 to get -3.

$$-6-3x-3\left(x-2\right)\left(x+2\right)=3x+6-1$$

Use the distributive property to multiply -3 by x-2.

$$-6-3x+\left(-3x+6\right)\left(x+2\right)=3x+6-1$$

Use the distributive property to multiply -3x+6 by x+2 and combine like terms.

$$-6-3x-3x^{2}+12=3x+6-1$$

Add -6 and 12 to get 6.

$$6-3x-3x^{2}=3x+6-1$$

Subtract 1 from 6 to get 5.

$$6-3x-3x^{2}=3x+5$$

Subtract 3x from both sides.

$$6-3x-3x^{2}-3x=5$$

Combine -3x and -3x to get -6x.

$$6-6x-3x^{2}=5$$

Subtract 6 from both sides.

$$-6x-3x^{2}=5-6$$

Subtract 6 from 5 to get -1.

$$-6x-3x^{2}=-1$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-3x^{2}-6x=-1$$

Divide both sides by -3.

$$\frac{-3x^{2}-6x}{-3}=-\frac{1}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\left(-\frac{6}{-3}\right)x=-\frac{1}{-3}$$

Divide -6 by -3.

$$x^{2}+2x=-\frac{1}{-3}$$

Divide -1 by -3.

$$x^{2}+2x=\frac{1}{3}$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+2x+1^{2}=\frac{1}{3}+1^{2}$$

Square 1.

$$x^{2}+2x+1=\frac{1}{3}+1$$

Add \frac{1}{3} to 1.

$$x^{2}+2x+1=\frac{4}{3}$$

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+1\right)^{2}=\frac{4}{3}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{\frac{4}{3}}$$

Simplify.

$$x+1=\frac{2\sqrt{3}}{3} x+1=-\frac{2\sqrt{3}}{3}$$

Subtract 1 from both sides of the equation.

$$x=\frac{2\sqrt{3}}{3}-1 x=-\frac{2\sqrt{3}}{3}-1$$